I have taken the function argument as char*, in my case I get XOR equal to 210. On the other side, In the other VS I took the first argument not char* but simply char[] and the XOR is 114, which is correct.
what's wrong? why can't I get the same thing here?
thanks for your replies.
UPDATE: You are right. sprintf() works fine. The problem is the rest of the code.
bool BuildAnglePacket(char* WR_PacketAZAngle, float AZAngle)
{
WR_PacketAZAngle[0] = 0x04;
WR_PacketAZAngle[1] = 0x30;
WR_PacketAZAngle[2] = 0x31;
WR_PacketAZAngle[3] = 0x02;
WR_PacketAZAngle[4] = 0x79;
WR_PacketAZAngle[5] = 0x4E;
WR_PacketAZAngle[6] = 0x48;
int XOR;
char HAnlge[9];
int iAzimuthAngle;
// AZAngle = 22;
if (AZAngle >= -22.5 && AZAngle <= 22.5)
{
iAzimuthAngle = AZAngle*10;
if( AZAngle < 0)
{
iAzimuthAngle= abs(iAzimuthAngle);
iAzimuthAngle=((~iAzimuthAngle)&0xFFFF) + 1 ;
}
iAzimuthAngle = 65536 + iAzimuthAngle;
sprintf(HAnlge,"%08X", iAzimuthAngle);
WR_PacketAZAngle[7] = HAnlge[0];
WR_PacketAZAngle[8] = HAnlge[1];
WR_PacketAZAngle[9] = HAnlge[2];
WR_PacketAZAngle[10] = HAnlge[3];
WR_PacketAZAngle[11] = HAnlge[4];
WR_PacketAZAngle[12] = HAnlge[5];
WR_PacketAZAngle[13] = HAnlge[6];
WR_PacketAZAngle[14] = HAnlge[7];
WR_PacketAZAngle[15] = 0x03;
for(int i=4;i<16;i++)
XOR ^= WR_PacketAZAngle[i];
WR_PacketAZAngle[16] = XOR;
WR_PacketAZAngle[17] ='\x0';
}
return true;
}
Solved: Yes, I forgot to initialize XOR.
Your problem is not with sprintf_s or sprintf. Value 65536 + 150 => 65686 => 0x10096.
This is the correct result as printed by your code, anything else would be a bug. BTW, I think you meant 150, not 15, as 0x96 => 150.
can it be that iAngle in your Windows version is unsigned short, so it wraps around and you actually get 150 instead of 65536+150? This would explain output of '00000096' but it means that it is a bug in the original calculation code, not with printing itself.
BTW, I assume that real code doesn't have char HAngle; but something like char HAngle[..]; otherwise anything can happen in case that compiler for some reason falls asleep and doesn't produce an error.
EDIT:
The updated code shows that XOR is not initialized and it can contain anything before it is used in calculation, so you can get any result. You have to set it to 0 first. On Windows side it probably worked if you were testing the debug version which sets integral variables to 0 or by pure chance.
The original question doesn't make much sense: the code won't compile unless you change HAngle to an array (or pointer to an array); hex 96 is decimal 150, not 15; and you get the extra 1 because you're adding 65536 (hex 10000) to the value.
In the updated question, you get an indeterminate value for XOR (and technically undefined behaviour) since you never initialise it. You want:
int XOR = 0;
^^^
int iAngle = 15; // for example
char HAngle;
iAngle = 65536 + iAngle;
So iAngle == 65536 + 15 == 65551 which in hex is 0x0001000F. If you printed that to a string, shouldn't you get the following?
[0] 48 '0'
[1] 48 '0'
[2] 48 '0'
[3] 49 '1'
[4] 48 '0'
[5] 48 '0'
[6] 48 '0'
[7] 70 'F'
[8] 0 '\0'
Surely index 3 must always be a hex 1 in this case?
It looks like the windows function is doing something strange...
Also HAngle should be an array if you're going to print to it, otherwise you're way overflowing the storage allocated to it. You seem to be able to treat it as a pointer when you call sprintf so I'm assuming you mean char *HAngle and have allocated memory for the buffer before you print to it?
EDIT: From your updated code it looks like XOR is not initialised? If it isn't then it could start with any random value (compiler doesn't have to set it to zero I'm afraid:) ). This could account for the different results. On both sides it could have an arbitrary initial value and just so happens on one side to have a zero initial value...
Related
// check if temp < 10
if (temp < 10) {
hexaDeciNum[i] = temp + 48;
i++;
}
else {
hexaDeciNum[i] = temp + 55;
i++;
}
n = n / 16;
}
I found this code to convert from decimal to hex but as you can see we have + 48 and + 55 anyone know why did we use these numbers? btw temp is to store the remainder... thanks!
What the code is doing, badly, is converting a value in the range of 0 to 15 into a corresponding character for the hexadecimal representation of that value. The right way to do that is with a lookup table:
const char hex[] = "0123456789ABCDEF";
hexaDeciNum[i] = hex[temp];
One problem with the code as written is that it assumes, without saying so, that you want the output characters encoded in ASCII. That's almost always the case, but there is no need to make that assumption. The compiler knows what encoding the system that it is targeting uses, so the values in the array hex in my code will be correct for the target system, even if it doesn't use ASCII.
Another problem with the code as written is the magic numbers. They don't tell you what their purpose is. To get rid of the magic numbers, replace 48 with '0' and replace 55 with 'A' - 10. But see the previous paragraph.
In C and C++ you can convert a base-10 digit to its corresponding character by adding it to '0', so
hexaDeciNum[i] = digit + '0';
will work correctly. There is no such requirement for any other values, so that conversion to a letter is not guaranteed to work, even if you use 'A' instead of that hardcoded 65.
And don't get me started on pointless comments:
// check if temp < 10
if (temp < 10)
If you look on the ASCII table you will see that the characters for numbers 0..9 are shifted by 48. So, if you take a number e.g. 0 and add 48 to it you will get a character for that number "0".
The same goes for characters if you take number 10 and add 55 to it you will get an "A" character from the ASCII table.
I have a c++ program that takes an integer and convert it to lower and uppercase alphabets, similar to what excel does to convert column index to column number but also including lower case letters.
#include <string>
#include <iostream>
#include <climits>
using namespace std;
string ConvertNum(unsigned long v)
{
char const digits[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
size_t const base = sizeof(digits) - 1;
char result[sizeof(unsigned long)*CHAR_BIT + 1];
char* current = result + sizeof(result);
*--current = '\0';
while (v != 0) {
v--;
*--current = digits[v % base];
v /= base;
}
return current;
}
// for testing
int main()
{
cout<< ConvertNum(705);
return 0;
}
I need the vba function to reverse this back to the original number. I do not have a lot of experience with C++ so I can not figure out a logic to reverse this in vba. Can anyone please help.
Update 1: I don't need already written code, just some help in the logic to reverse it. I'll try to convert the logic into code myself.
Update 2: Base on the wonderful explanation and help provided in the answer, it's clear that the code is not converting the number to a usual base52, it is misleading. So I have changed the function name to eliminate the confusion for future readers.
EDIT: The character string format being translated to decimal by the code described below is NOT a standard base-52 schema. The schema does not include 0 or any other digits. Therefore this code should not be used, as is, to translate a standard base-52 value to decimal.
O.K. this is based on converting a single character based on its position in a long string. the string is:
chSET = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
The InStr() function tells us the A is in position 1 and the Z is in position 26 and that a is in position 27. All characters get converted the same way.
I use this rather than Asc() because Asc() has a gap between the upper and lower case letters.
The least significant character's value gets multiplied by 52^0The next character's value gets multiplied by 52^1The third character's value gets multiplied by 52^3, etc. The code:
Public Function deccimal(s As String) As Long
Dim chSET As String, arr(1 To 52) As String
Dim L As Long, i As Long, K As Long, CH As String
chSET = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
deccimal = 0
L = Len(s)
K = 0
For i = L To 1 Step -1
CH = Mid(s, i, 1)
deccimal = deccimal + InStr(1, chSET, CH) * (52 ^ K)
K = K + 1
Next i
End Function
Some examples:
NOTE:
This is NOT the way bases are usually encoded. Usually bases start with a 0 and allow 0 in any of the encoded value's positions. In all my previous UDF()'s similar to this one, the first character in chSET is a 0 and I have to use (InStr(1, chSET, CH) - 1) * (52 ^ K)
Gary's Student provided a good and easy to understand way to get the number from what I call "Excel style base 52" and this is what you wanted.
However this is a little different from the usual base 52. I'll try to explain the difference to regular base 52 and its conversion. There might be an easier way but this is the best I could come up with that also explains the code you provided.
As an example: The number zz..zz means 51*(1 + 52 + 52^2 + ... 52^(n-1)) in regular base 52 and 52*(1 + 52 + 52^2 + ... 52^(n-1)) in Excel style base 52. So Excel style get's higher number with fewer digits. Here is how much that difference is based on number of digits. How is this possible? It uses leading zeros so 1, 01, 001 etc are all different numbers. Why don't we do this normally? It would mess up the easy arithmetic of the usual system.
We can't just shift all the digits by one after the base change and we can't just substract 1 before the base change to counter the fact that we start at 1 instead of 0. I'll outline the problem with base 10. If we'd use Excel style base 10 to number the columns, we would have to count like "0, 1, 2, ..., 9, 00, 01, 02, ...". On the first glance it looks like we just have to shift the digits so we start counting at 1 but this only works up to the 10th number.
1 2 .. 10 11 .. 20 21 .. 99 100 .. 110 111 //normal counting
0 1 .. 9 00 .. 09 10 .. 88 89 .. 99 000 //excel style counting
You notice that whenever we add a new digit we shift again. To counter that, we have to do a shift by 1 before calculating each digit, not shift the digit after calculating it. (This only makes a difference if we're at 52^k) Note that we still assign A to 0, B to 1 etc.
Normally what you would do to change bases is looping with something like
nextDigit = x mod base //determining the last digit
x = x/base //removing the last digit
//terminate if x = 0
However now it is
x = x - 1
nextDigit = x mod base
x = x/base
//terminate if x = 0
So x is decremented by 1 first! Let's do a quick check for x=52:
Regular base 52:
nextDigit = x mod 52 //52 mod 52 = 0 so the next digit is A
x = x/52 //x is now 1
//next iteration
nextDigit = x mod 52 //1 mod 52 = 1 so the next digit is B
x = x/52 //1/52 = 0 in integer arithmetic
//terminate because x = 0
//result is BA
Excel style:
x = x-1 //x is now 51
nextDigit = x mod 52 //51 mod 52 = 51 so the next digit is z
x = x/52 //51/52 = 0 in integer arithmetic
//terminate because x=0
//result is z
It works!
Part 2: Your C++ code
Now for let's read your code:
x % y means x mod y
When you do calculations with integers, the result will be an integer which is achieved by rounding down. So 39/10 will produce 3 etc.
x++ and ++x both increment x by 1.
You can use this in other statements to save a line of code. x++ means x is incremented after the statement is evaluated and ++x means it is incremented before the statement is evaluated
y=f(x++);
is the same as
y = f(x);
x = x + 1;
while
y=f(++x);
is the same as
x = x + 1;
y = f(x);
This goes the same way for --
Char* p creates a pointer to a char.
A pointer points to a certain location in memory. If you change the pointer, it points to a different location. E.g. doing p-- moves the pointer one to the left. To read or write the value that is saved at the location, use *p. E.g. *p="a"; "a" is written to the memory location that p points at. *p--="a"; "a" is written to the memory but the pointer is moved to the left afterwards so *p is now whatever is in the memory left of "a".
strings are just arrays of type char.
The end of a string is always '\0' if the computer reads a string it continues until it finds '\0'
This is hopefully enough to understand the code. Here it is
#include <string>
#include <iostream>
#include <climits>
using namespace std;
string base52(unsigned long v)
{
char const digits[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"; //The digits. (Arrays start at 0)
size_t const base = sizeof(digits) - 1; //The base, based on the digits that were given
char result[sizeof(unsigned long)*CHAR_BIT + 1]; //The array that holds the answer
//sizeof(unsigned long)*CHAR_BIT is the number of bits of an unsigned long
//which means it is the absolute longest that v can be in any base.
//The +1 is to hold the terminating character '\0'
char* current = result + sizeof(result); //This is a pointer that is supposed to point to the next digit. It points to the first byte after the result array (because its start + length)
//(i.e. it will go through the memory from high to low)
*--current = '\0'; //The pointer gets moved one to the left (to the last char of result and the terminating char is added
//the pointer has to be moved to the left first because it was actually pointing to the first byte after the result.
while (v != 0) { //loop until v is zero (until there are no more digits left.
v--; //v = v - 1. This is the important part that does the 1 -> A part
*--current = digits[v % base]; // the pointer is moved one to the left and the corresponding digit is saved
v /= base; //the last digit is dropped
}
return current; //current is returned, which points at the last saved digit. The rest of the result array (before current) is not used.
}
// for testing
int main()
{
cout<< base52(705);
return 0;
}
Brian Kernighnan in his book Programming with C says
By definition, chars are just small integers, so char variables and
constants are identical to ints in arithmetic expressions.
Does this mean we can subtract char variable from int ??
I wrote a small piece of code:
#include <stdio.h>
main()
{
int a ;
int c;
a = 1;
c = 1 - '0' ;
printf("%d", c);
}
However it gives me output = -47...
What is that I'm doing wrong ?? Are the variables I assigned have the right type??
The output is to be expected. '0' is a char value that, since your compiler presumably uses the ASCII encoding, has value 48. This is converted to int and subtracted from 1. Which gives the value -47.
So the program does what it is expected to do, just not what you might hope it would do. As for what you are doing wrong, it is hard to say. I'm not sure what you expect the program to do, or what problem you are trying to solve.
The characters from '0'-'9'' have values 48-57 when converted to integer ('0' = 48, '1' = 49 etc). Read more about ASCII Values. When used in numerical calculation, first they are converted to int, so 1- '0' = 1-48 =-47.
You're mixing here the actual operation with the form of representation. printf outputs the data according to the specified format - integer in your case. If you want to print it as a character, switch %d with %c.
What you are doing is treating with the ASCII code of the chars, each char has an ASCII value assigned.
Now, playing a little with the ASCII of each char you can do things like:
int a;
a = 'a' - 'A' ;
printf("%d", a);
And get 32 as output, due to the ASCII value to 'a' = 97 and for 'A' = 65, then you have 97-65 = 32
I think this gives you more clear understanding...
#include <stdio.h>
main()
{
int a ;
a = 1;
printf("%d", char(a+48));
//or printf("%d", char(a+'0'));
}
Im able to convert most things without a problem, a google search if needed. I cannot figure this one out, though.
I have a char array like:
char map[40] = {0,0,0,0,0,1,1,0,0,0,1,0,1... etc
I am trying to convert the char to the correct integer, but no matter what I try, I get the ascii value: 48/ 49.
I've tried quite a few different combinations of conversions and casts, but I cannot end up with a 0 or a 1, even as a char.
Can anyone help me out with this?
Thanks.
The ascii range of the characters representing integers is 48 to 57 (for '0' to '9'). You should subtract the base value 48 from the character to get its integer value.
char map[40] = {'0','0','0','0','0','1','1','0','0','0','1','0','1'...};
int integerMap[40];
for ( int i = 0 ;i < 40; i++)
{
integerMap[i] = map[i] - 48 ;
// OR
//integerMap[i] = map[i] - '0';
}
If the char is a literal, e.g. '0' (note the quotes), to convert to an int you'd have to do:
int i = map[0] - '0';
And of course a similar operation across your map array. It would also be prudent to error-check so you know the resulting int is in the range 0-9.
The reason you're getting 48/49 is because, as you noted, direct conversion of a literal like int i = (int)map[0]; gives the ASCII value of the char.
I have not many experience with operations/storage of binary data so I would greatly appreciate if someone could clarify some things for me.
I have a device say where you have to store 16 bytes. e.g., you should send it an array of bytes proceeded probably with header information. e.g., smth like this:
unsigned char sendBuffer[255];
sendBuffer[0] = headerInfo1;
sendBuffer[1] = headerInfo1;
sendBuffer[2] = headerInfo1;
sendBuffer[3] = headerInfo1;
sendBuffer[4] = data;
sendBuffer[5] = data;
sendBuffer[6] = data;
sendBuffer[7] = data;
sendBuffer[8] = data;
...
sendBuffer[20] = data;
Let's say send operation is easy, you just use Send(sendBuffer, length).
My question is say I want to store an integer in the device - what is the best way to do this?
I have a sample code which does it and I was not sure if it was ok and how it was doing it. It confused me too. I basically enter the number I want to store in text box. Say I want to store 105 in decimal. I enter "00000000000000000000000000000105" (I am not sure how program interprets this yet, as decimal or as hex), then there is this code:
for(int i=0,m=0; i < size; i+=2,m++)
{
char ch1, ch2;
ch1=(char)str[i]; // str contains the number I entered above as string, padded
ch2=(char)str[i+1];
int dig1, dig2;
if(isdigit(ch1)) dig1 = ch1 - '0';
else if(ch1>='A' && ch1<='F') dig1 = ch1 - 'A' + 10;
else if(ch1>='a' && ch1<='f') dig1 = ch1 - 'a' + 10;
if(isdigit(ch2)) dig2 = ch2 - '0';
else if(ch2>='A' && ch2<='F') dig2 = ch2 - 'A' + 10;
else if(ch2>='a' && ch2<='f') dig2 = ch2 - 'a' + 10;
// Contains data to write as a byte array; this is basically the 'data' part as mentioned in my above snippet
array1[m] = (char)(dig1*16 + dig2);
}
And this array1[m] is written to the device using Send as above. But when I debug array1 contains: 0000000000000015
When I do the read the value I get is correct, it is 00000000000000000000000000000105. How come this works?
You're reinventig a few wheels here, but that's to be expected if you're new to C++.
std::cin >> yourInteger will read an integer, no need to convert that yourself.
Leading zeroes are usually not written out, but in a C++ integer type they're always present. E.g. int32_t always has 32 bits. If it stores 105 (0x69), it really stores 0x00000069.
So, the best way is probably to memcpy that integer to your sendBuffer. You should copy sizeof(yourInteger) bytes.
Seems there are a few questions hiding in here, so some extra answers:
You say that array1 contains: 0000000000000015, not 105.
Well, it's an array, and each member is shown as an 8 bits integer in its own right.
E.g. the last value is 5 or 05, that's the same after all. Similarly, the penultimate integer is 1 or 01.
You also wrote "Say I want to store 105 in decimal. I enter 00000000000000000000000000000105". That doesn't actually store 105 decimal. It stores 105 hexadecimal, which is 261 decimal. It is the string to integer conversion which determines the final value. If you would use base 18 (octodecimal), the string "105" becomes the integer 1*18*18 + 0 + 5 = 329 (decimal), and that would be stored as 000000101001001 binary.